NCERT Solutions for Class 12 Chemistry Chapter 1 โ Solutions (English Medium)
๐ Get NCERT Solutions for Class 12 Chemistry Chapter 1 โ Solutions with accurate answers and step-by-step explanations in a clean, exam-friendly format. This chapter covers types of solutions (solid, liquid, gaseous), concentration terms (mass %, mole fraction, molarity, molality), solubility, Raoultโs law, colligative properties (ฮTb, ฮTf, osmotic pressure), vanโt Hoff factor, and abnormal molar mass. Perfect for CBSE board exams and NEET/JEE practice. โ
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Q.1: Define the true solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer
โ
True solution
is a
homogeneous mixture
of
two or more chemically non-reacting substances
(solute +
solvent).๐ In a true solution, solute particles are extremely small (generally < 1 nm) and are uniformly distributed , so the mixture looks the same throughout.
โ Types of solutions formed: 9 types (based on the physical state of solute and solvent).
Explanation (Step by Step)
๐ค A solution can be classified by what is dissolved
(solute)
and in what it is dissolved
(solvent).So we get 3 solvent states (Gas, Liquid, Solid) ร 3 solute states (Gas, Liquid, Solid) = 9 types .
A) Gaseous Solutions (Solvent = Gas)
1) Gas in Gas
๐ Example: Air (O2 + N2 + CO2 etc.)
๐ Example: Air (O2 + N2 + CO2 etc.)
2) Liquid in Gas
๐ Example: Water vapour in air (humidity)
๐ Example: Water vapour in air (humidity)
3) Solid in Gas
๐ Example: Smoke/dust in air, or camphor vapours in N2
๐ Example: Smoke/dust in air, or camphor vapours in N2
B) Liquid Solutions (Solvent = Liquid)
4) Gas in Liquid
๐ Example: CO2 dissolved in water (aerated drinks), O2 dissolved in water (needed by aquatic life)
๐ Example: CO2 dissolved in water (aerated drinks), O2 dissolved in water (needed by aquatic life)
5) Liquid in Liquid
๐ Example: Ethanol in water (alcohol + water mixture)
๐ Example: Ethanol in water (alcohol + water mixture)
6) Solid in Liquid
๐ Example: Sugar in water, NaCl in water (saline water, concentration often written as g/L or mol/L)
๐ Example: Sugar in water, NaCl in water (saline water, concentration often written as g/L or mol/L)
C) Solid Solutions (Solvent = Solid)
7) Gas in Solid
๐ Example: Hydrogen in palladium (H2 absorbed in Pd)
๐ Example: Hydrogen in palladium (H2 absorbed in Pd)
8) Liquid in Solid
๐ Example: Amalgams like NaโHg (sodium amalgam)
๐ Example: Amalgams like NaโHg (sodium amalgam)
9) Solid in Solid
๐ Example: Alloys such as gold ornaments (Au mixed with Cu/Ag), brass (Cu + Zn)
๐ Example: Alloys such as gold ornaments (Au mixed with Cu/Ag), brass (Cu + Zn)
Did You Know?
โ
True solutions do not show
Tyndall effect
(they donโt scatter light), so a beam of light is not visible through them.โ The concentration of a solution is commonly expressed as mol/L (molarity) , g/L , or % (w/v, v/v) depending on the need.
Q.2: Give an example of a solid solution in which the solute is a gas.
Answer
โ
๐
Hydrogen gas dissolved in palladium
is a solid solution where the
solute = gas (H2)
and the
solvent = solid (Pd)
.
Explanation (Step by Step)
๐ Some metals (like
palladium
)
can absorb gases into their
crystal lattice
.๐ค So, H2 (gas) gets uniformly distributed inside Pd (solid) , forming a gas-in-solid solution .
Did You Know?
โ
Palladium can absorb a
surprisingly large amount of hydrogen
,
which is why itโs studied in
hydrogen storage research
.
Q.3: Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Answer
(i) Mole fraction (ฯ)
๐ Mole fraction is the ratio of moles of a component to the total moles of all components in the solution.
๐ For a binary solution (A + B):
โข ฯA = nA / (nA + nB)
โข ฯB = nB / (nA + nB)
โ ฯA + ฯB = 1
๐ค Unit: No unit (dimensionless)
๐ Mole fraction is the ratio of moles of a component to the total moles of all components in the solution.
๐ For a binary solution (A + B):
โข ฯA = nA / (nA + nB)
โข ฯB = nB / (nA + nB)
โ ฯA + ฯB = 1
๐ค Unit: No unit (dimensionless)
(ii) Molality (m)
๐ Molality is the number of moles of solute present in 1 kg of solvent .
๐ Formula:
โข m = (Moles of solute) / (Mass of solvent in kg)
โข m = nsolute / wsolvent(kg)
โ Unit: mol kgโ1 (or mol/kg)
๐ Molality is the number of moles of solute present in 1 kg of solvent .
๐ Formula:
โข m = (Moles of solute) / (Mass of solvent in kg)
โข m = nsolute / wsolvent(kg)
โ Unit: mol kgโ1 (or mol/kg)
(iii) Molarity (M)
๐ Molarity is the number of moles of solute present in 1 litre (1 L) of solution .
๐ Formula:
โข M = (Moles of solute) / (Volume of solution in L)
โข M = nsolute / Vsolution(L)
โ Unit: mol Lโ1 (or mol/L)
๐ Molarity is the number of moles of solute present in 1 litre (1 L) of solution .
๐ Formula:
โข M = (Moles of solute) / (Volume of solution in L)
โข M = nsolute / Vsolution(L)
โ Unit: mol Lโ1 (or mol/L)
(iv) Mass percentage (w/w %)
๐ Mass percentage of a component is the mass of that component in the solution divided by the total mass of solution, multiplied by 100 .
๐ For solute A in solvent B:
โข Mass % (A) = [ wA / (wA + wB) ] ร 100
โ Unit: % (percentage)
๐ Mass percentage of a component is the mass of that component in the solution divided by the total mass of solution, multiplied by 100 .
๐ For solute A in solvent B:
โข Mass % (A) = [ wA / (wA + wB) ] ร 100
โ Unit: % (percentage)
Explanation (Step by Step)
๐ค Many students confuse these terms, so remember this simple idea:๐ Mole fraction โ compares moles with total moles (no unit)
๐ Molality โ moles per kg of solvent (mol/kg), temperature independent
๐ Molarity โ moles per L of solution (mol/L), changes with temperature (because volume changes)
๐ Mass % โ compares mass with total mass (in %)
๐ Extra useful form (when masses are given):
โข n = w(g) / M(g molโ1)
So molality can also be written as:
โข m = wsolute(g) / [ Msolute(g molโ1) ร wsolvent(kg) ]
Did You Know?
โ
Molality (m)
is preferred in many lab calculations because it does not change with temperature, while
molarity (M)
can change when the solution expands or contracts.โ Mole fraction is super useful in topics like Raoultโs law and vapour pressure of solutions because it directly connects with gaseous behavior.
Q.4: Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution.
What should be the molarity of such a sample if the density of the solution is 1.504 g mLโ1?
Answer
โ
๐ Molarity of
68% (w/w) HNO3 solution
=
16.23 mol Lโ1
(โ
16.2 M
)
Explanation (Step by Step)
๐ Given
โข 68% (w/w) HNO3 means 68 g HNO3 in 100 g solution
โข Density (ฯ) = 1.504 g mLโ1
โข Molar mass of HNO3 = (1) + (14) + (3 ร 16) = 63 g molโ1
โข 68% (w/w) HNO3 means 68 g HNO3 in 100 g solution
โข Density (ฯ) = 1.504 g mLโ1
โข Molar mass of HNO3 = (1) + (14) + (3 ร 16) = 63 g molโ1
Step 1: Find moles of HNO3 ๐ค
๐ n = w / M
n = 68 g / 63 g molโ1
n = 1.079 mol (approx)
๐ n = w / M
n = 68 g / 63 g molโ1
n = 1.079 mol (approx)
Step 2: Find volume of 100 g solution (using density) ๐
๐ Volume = Mass / Density
V = 100 g / 1.504 g mLโ1
V = 66.5 mL
Convert to litres:
66.5 mL = 0.0665 L
๐ Volume = Mass / Density
V = 100 g / 1.504 g mLโ1
V = 66.5 mL
Convert to litres:
66.5 mL = 0.0665 L
Step 3: Calculate molarity (M) โ
๐ M = n / V(L)
M = 1.079 mol / 0.0665 L
M = 16.23 mol Lโ1
โ Therefore, Molarity = 16.23 mol Lโ1 (โ 16.2 M )
๐ M = n / V(L)
M = 1.079 mol / 0.0665 L
M = 16.23 mol Lโ1
โ Therefore, Molarity = 16.23 mol Lโ1 (โ 16.2 M )
Did You Know?
โ
โConcentratedโ acids in labs are often given in
% (w/w)
,
not molarity.๐ To convert % (w/w) โ molarity , density is essential because volume changes with concentration.
๐ Thatโs why the same acid can have different molarity if its density (or temperature) changes.
Q.5: A solution of glucose in water is labelled as 10% w/w. What would be the molality and mole fraction of each component in the solution?
If the density of solution is 1.2 g mLโ1, then what shall be the molarity of the solution?
Answer
โ
๐ For
10% (w/w) glucose solution
:โข Molality (m) = 0.61 mol kgโ1
โข Mole fraction of glucose (ฯglucose) โ 0.01
โข Mole fraction of water (ฯwater) โ 0.99
โข Molarity (M) (given density 1.2 g mLโ1 ) = 0.666 mol Lโ1 (โ 0.67 M )
Explanation (Step by Step)
๐ Meaning of
10% w/w
:
10% (w/w) glucose โ 10 g glucose present in 100 g solution.
So water = 100 โ 10 = 90 g
10% (w/w) glucose โ 10 g glucose present in 100 g solution.
So water = 100 โ 10 = 90 g
Given ๐
โข Glucose = C6H12O6 ,
โข molar mass = (6ร12) + (12ร1) + (6ร16) = 180 g molโ1
โข Water = H2O , molar mass = (2ร1) + 16 = 18 g molโ1
โข Density of solution = 1.2 g mLโ1
โข Glucose = C6H12O6 ,
โข molar mass = (6ร12) + (12ร1) + (6ร16) = 180 g molโ1
โข Water = H2O , molar mass = (2ร1) + 16 = 18 g molโ1
โข Density of solution = 1.2 g mLโ1
(i) Molality (m) ๐ค
๐ Molality = (moles of solute) / (mass of solvent in kg)
โข Moles of glucose = 10 g / 180 g molโ1 = 0.0556 mol
โข Mass of water = 90 g = 0.090 kg
So,
m = 0.0556 mol / 0.090 kg = 0.617 mol kgโ1
โ m โ 0.61 mol kgโ1
๐ Molality = (moles of solute) / (mass of solvent in kg)
โข Moles of glucose = 10 g / 180 g molโ1 = 0.0556 mol
โข Mass of water = 90 g = 0.090 kg
So,
m = 0.0556 mol / 0.090 kg = 0.617 mol kgโ1
โ m โ 0.61 mol kgโ1
(ii) Mole fraction of each component ๐
๐ First find moles:
โข nglucose = 10/180 = 0.0556 mol
โข nwater = 90/18 = 5.00 mol
Total moles = 0.0556 + 5.00 = 5.0556 mol
โ Mole fraction of glucose:
ฯglucose = 0.0556 / 5.0556 = 0.011 โ 0.01
โ Mole fraction of water:
ฯwater = 5.00 / 5.0556 = 0.989 โ 0.99
(Or simply: ฯwater = 1 โ ฯglucose )
๐ First find moles:
โข nglucose = 10/180 = 0.0556 mol
โข nwater = 90/18 = 5.00 mol
Total moles = 0.0556 + 5.00 = 5.0556 mol
โ Mole fraction of glucose:
ฯglucose = 0.0556 / 5.0556 = 0.011 โ 0.01
โ Mole fraction of water:
ฯwater = 5.00 / 5.0556 = 0.989 โ 0.99
(Or simply: ฯwater = 1 โ ฯglucose )
(iii) Molarity (M) using density โ
๐ค We need volume of solution for 100 g solution .
๐ Volume = Mass / Density
V = 100 g / 1.2 g mLโ1 = 83.33 mL = 0.08333 L
Now,
M = moles of glucose / volume of solution (L)
M = 0.0556 mol / 0.08333 L = 0.667 mol Lโ1
โ M โ 0.666 mol Lโ1 (โ 0.67 M )
๐ค We need volume of solution for 100 g solution .
๐ Volume = Mass / Density
V = 100 g / 1.2 g mLโ1 = 83.33 mL = 0.08333 L
Now,
M = moles of glucose / volume of solution (L)
M = 0.0556 mol / 0.08333 L = 0.667 mol Lโ1
โ M โ 0.666 mol Lโ1 (โ 0.67 M )
Did You Know?
โ
Molality depends on
mass of solvent
,
so it stays almost constant even if temperature changes.โ Molarity depends on volume of solution , so it can change with temperature (volume expands/contracts).
๐ Thatโs why density helps a lot when converting % w/w โ molarity .